If A is a symmetric matrixfand `n in N`, then `A^(n)` is

Options

A. symmetric matrix
B. skew-symmetric matrix
C. a diagonal matrix
D. identiy matrix

Answer

To solve the problem, we need to determine the nature of the matrix \( A^n \) when \( A \) is a symmetric matrix and \( n \) is a natural number.

### Step-by-Step Solution:

1. **Definition of Symmetric Matrix**:
A matrix \( A \) is symmetric if \( A = A^T \), where \( A^T \) is the transpose of \( A \).

2. **Taking the Transpose of \( A^n \)**:
We need to find the transpose of \( A^n \). By the property of transposes, we have:
\[ (A^n)^T = (A \cdot A \cdot A \cdots A)^T = A^T \cdot A^T \cdots A^T \quad (n \text{ times}) \]

3. **Using the Symmetric Property**:
Since \( A \) is symmetric, we know that \( A^T = A \). Therefore, we can substitute \( A \) for \( A^T \):
\[ (A^n)^T = A \cdot A \cdots A = A^n \]

4. **Conclusion**:
Since \( (A^n)^T = A^n \), we conclude that \( A^n \) is also symmetric.

### Final Answer:
Thus, if \( A \) is a symmetric matrix and \( n \) is a natural number, then \( A^n \) is a symmetric matrix.